Question

Show that (3.9) follows from (3.8) . Hint: Give a proof by contradiction. Let ${\mathit{S}}_{\mathbf{\alpha }\mathbf{\beta }}$be the parenthesis in ; you may find it useful to think of the components written as a matrix. You want to prove that all 9 components of ${\mathit{S}}_{\mathbf{\alpha \beta }}$are zero. Suppose it is claimed that ${\mathit{S}}_{\mathbf{12}}$is not zero. Since $\mathit{V}{\mathbf{\text{'}}}_{\mathbf{\beta }}$is an arbitrary vector, take it to be the vector , and observe that ${\mathit{S}}_{\mathbf{\alpha \beta }}\mathit{V}{\mathbf{\text{'}}}_{\mathbf{\beta }}$is then not zero in contradiction to .Similarly show that all components of ${\mathit{S}}_{\mathbf{\alpha \beta }}$are zero as claims.

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The vector V and S is a tensor.

Definition of a cartesian tensor.

The first rank tensor is just a vector. A tensor of the second rank has nine components (in three dimensions) in every rectangular coordinate system.

Prove the statement.

Let The vector V and S is a tensor.

The formula states that $\left(T{\text{'}}_{\alpha \beta }-a_{\alpha i}{a}_{\beta i}{T}_{ij}\right)V{\text{'}}_{\beta }=0$.

For any $\alpha =0,S_{\alpha \beta }V{\text{'}}_{\beta }=0$

$$\begin{gathered} SV\text{'}=\left(\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right)\left(\begin{array}{c}V{\text{'}}_1\\ V{\text{'}}_2\\ V{\text{'}}_3\end{array}\right) \\ SV\text{'}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

Let S be the vectors, rows of s are denoted by $S_i=\left(S_{i1},S_{i2},S_{i3}\right)$